Question

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in E:\

<?
$link = mysql_connect("localhost","root","000000");
mysql_query("use example1");
$sqlstr="select * from member1 where name ='".$_POST['member1_name']."' and passwd = '".$_POST['member1_passwd']."'";
$result = mysql_query($sqlstr,$link);
if (mysql_num_rows ($result)==1){
echo "<center>会员成功登陆</center>";
}else{
echo "<center>会员登录失败</center>";
}
mysql_free_result($result);
mysql_close($link);
?>

Answer 1

你改成这样吧：
<?
$link = mysql_connect("localhost","root","000000");
mysql_select_db("use example1");//不是mysql_query('use ...');...
$sqlstr="select * from member1 where name ='$_POST[member1_name]' and passwd = '$_POST[member1_passwd]'";//$_POST[member1_name]和$_POST['member1_name']一样的，在速度不一样，不过没关系啦。
$result = mysql_query($sqlstr);//直接执行执行就好了。
if (mysql_num_rows ($result)==1){
echo "<center>会员成功登陆</center>";
}else{
echo "<center>会员登录失败</center>";
}
mysql_free_result($result);//你出错就在这里。那你干脆别要了。
mysql_close($link);//你八成教科书看多了。呵呵，最后这两行要不要无所谓。
?>

Answer 2

Connecting to 3.22, 3.23 & 4.0 servers is not。。。后面是啥啊`关健部分给你截断了`哈哈
是php警告你了：mysql_connect()连接数据库时有问题了
看看你的名字密码都对不`还有路径`会不会是找不到`没指对